Calculating Odd/Even Number of Divisors in Python

The challenge

Given an integer n return "odd" if the number of its divisors is odd. Otherwise, return "even".

Note: big inputs will be tested.

Examples:

All prime numbers have exactly two divisors (hence "even").

For n = 12 the divisors are [1, 2, 3, 4, 6, 12] – "even".

For n = 4 the divisors are [1, 2, 4] – "odd".

The solution in Python code

Option 1:

def oddity(n):
    #your code here
    return 'odd' if n**0.5 == int(n**0.5) else 'even'

Option 2:

import math
def oddity(n):
    return math.sqrt(n) % 1 == 0 and 'odd' or 'even'

Option 3:

oddity=lambda n: ["odd","even"][n**.5%1!=0]

Test cases to validate our solution

import test
from solution import oddity

@test.describe("Sample tests")
def tests():
    @test.it("Some examples")
    def tests():
        test.assert_equals(oddity(1), 'odd')
        test.assert_equals(oddity(5), 'even')
        test.assert_equals(oddity(16), 'odd')